The correct order of the following complexes in terms of their crystal field stabilization energies is :

The correct order of the following complexes in terms of their crystal field stabilization energies is :
  1. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt \left[\mathrm{Co}(\mathrm{en})_3\right]^{3+} \lt$ $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$
  2. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+} \lt$ $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$
  3. $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt$ $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}$
  4. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+} \lt \left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+} \lt$ $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$

Solution

Crystal field splitting energy $(\Delta) \propto$ charge or oxidation state of central metal atom.
Crystal field splitting energy $(\Delta) \propto$ Field strength of ligand (and chelation)
Crystal field stabilisation energy (CFSE) $=\left[-0.4 \mathrm{t}_{\mathrm{g}_g}\right.$ +0.6 eg] $\Delta_{\circ}$ (for octahedral)
For,
$\begin{aligned} & {\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}: \mathrm{Co}^{3+}: \mathrm{t}_{2 \mathrm{~g}}^6 \mathrm{eg}^0 ; \mathrm{CFSE}=-2.4\left(\Delta_0\right)_1} \\ & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}: \mathrm{Co}^{3+}: \mathrm{t}_{2 \mathrm{~g}}^6 \mathrm{eg}^0 ; \mathrm{CFSE}=-2.4\left(\Delta_0\right)_2} \\ & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{2+}: \mathrm{Co}^{2+}: \mathrm{t}_{2 g}^5 \mathrm{eg}^2 ; \mathrm{CFSE}=-0.8\left(\Delta_0\right)_3} \\ & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\right]^{2+}: \mathrm{Co}^{2+}: \mathrm{e}^4 \mathrm{t}_2^3 ; \mathrm{CFSE}=-1.2 \Delta_{\mathrm{t}}} \\ & \left(\mathrm{as}:\left(\Delta_{\mathrm{t}}\right)=\frac{4}{9}\left(\Delta_0\right)_3\right) \\ & \because \quad \Delta_{\mathrm{t}} < \left(\Delta_0\right)_3 < \left(\Delta_0\right)_2 < \left(\Delta_0\right)_1\end{aligned}$

Asked in: JEE Main 2025 (22 Jan Shift 2)

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